Suppose that $\{\mu_k\}$ is a sequence of probability measures defined on $(I,\mathcal{B})$, where $I$ is a closed real interval and $\mathcal{B}$ is its Borel $\sigma$-algebra, such that:

  • each $\mu_k$ is made of countably many atoms $\{p^n_k\}$;
  • for every $k$, the set $\bigcup_{n\in\mathbb{N}}p^n_k$ is not dense in $I$;
  • $\bigcup_{n,k\in\mathbb{N}}p^n_k$ is dense in $I$.

I wonder which further assumptions are needed to ensure that:

  • $\mu_k$ weakly converges to a probability measure $\mu$,
  • $\mu$ is absolutely continuous.

Thanks a lot!

  • 2
    $\begingroup$ You would need to know something about the total mass in intervals. Otherwise every measure, continuous or otherwise is a limit of measures like you describe. $\endgroup$ – Anthony Quas Nov 5 '20 at 20:23
  • 1
    $\begingroup$ Is it true that you assume that $(\mu_k)$ is a convergent sequence where $\mu$ is an arbitrary probability measure? Otherwise there is no indication that $(\mu_k)$ converges. $\endgroup$ – Dieter Kadelka Nov 5 '20 at 21:26
  • $\begingroup$ Clarified, thanks. $\endgroup$ – Alessandro Della Corte Nov 5 '20 at 21:37
  • $\begingroup$ For the first question: I doubt if there is something better than the portemanteu theorem. Maybe you get nothing better than tautologies or special cases. $\endgroup$ – Dieter Kadelka Nov 5 '20 at 23:17
  • $\begingroup$ You might be looking for uniform absolute continuity. $\endgroup$ – Nate Eldredge Nov 15 '20 at 19:06

(I cannot make any comment so I have no choice but post my comment as answer)
I'm not sure which kind of $\mu_k$ you want to control. You are surely given tightness for free, so mainly if you can describe the limit law by some total set of linear functionals then your work is done.
I think you know there are a lot of possible choices like moment generating function, characteristic function, cumulative distribution function.
Or you may try to prove the convergence of moments, or of Stieljes transform.
Or, you may craft one yourself based on properties of your $\mu_k$.
Hope this help.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.